The tree diagram of the problem above is attached
There are four outcomes of the two events,
First test - Cancer, Second Test - Cancer, the probability is 0.0396
First test - Cancer, Second Test - No Cancer, the probability is 0.0004
First test - No Cancer, Second Test - There is cancer, the probability is 0.0096
First test - No cancer, Second Test - No cancer, the probability is 0.9054
The probability of someone picked at random has cancer given that test result indicates cancer is

The probability of someone picked at random has cancer given that test result indicates no cancer is
Answer:
x = 8
Step-by-step explanation:
Δ TSU and Δ TRV are similar , thus the ratios of corresponding sides are equal, that is
=
, substitute values
=
=
( cross- multiply )
3x = x + 16 ( subtract x from both sides )
2x = 16 ( divide both sides by 2 )
x = 8
If there were 49 cars in a line that stretched 528 feet, what is the average car length? assume that the cars are lined up bumper-to-bumper
Answer: We are given there are 49 cars
Also these 49 cars are in a line stretched 528 feet.
Now the average length of the car is:
Average length = 
=
Therefore, the average length of car is 10.78 feet
Answer:
13.5
Step-by-step explanation:
c = 2πr
42.3 = 2(3.14)r
42.3 = 6.28r
6.74 = r
d = r + r
d = 6.74 + 6.74
d = 13.5
Answer:
Given bivariate data, first determine which is the independent variable, x, and which is the dependent variable, y. Enter the data pairs into the regression calculator. Substitute the value for one variable into the equation for the regression line produced by the calculator, and then predict the value of the other variable.
Step-by-step explanation: